2015 AMC 10B Problem 5

P1-10~2 minPrint

Logic Puzzles Set up the equation/formula and compute; no special trick needed

David, Hikmet, Jack, Marta, Rand, and Todd were in a $12$ -person race with $6$ other people. Rand finished $6$ places ahead of Hikmet. Marta finished $1$ place behind Jack. David finished $2$ places behind Hikmet. Jack finished $2$ places behind Todd. Todd finished $1$ place behind Rand. Marta finished in $6$ th place. Who finished in $8$ th place?

Show solution without answering

Key insight

Only Marta's place is given as an exact number, 6th. Start there and follow the clues one by one: Jack 5th, Todd 3rd, Rand 2nd, Hikmet 8th.

Solution

Most clues only compare two runners. Only one clue gives an actual place: Marta is $6$ th. So start with Marta and follow the chain.

Remember: "ahead" means a smaller place number, and "behind" means a larger one.

Marta is $6$ th.
Marta is $1$ place behind Jack, so Jack is $6 - 1 = 5$ th.
Jack is $2$ places behind Todd, so Todd is $5 - 2 = 3$ rd.
Todd is $1$ place behind Rand, so Rand is $3 - 1 = 2$ nd.
Rand is $6$ places ahead of Hikmet, so Hikmet is $2 + 6 = 8$ th.
(For completeness: David is $2$ places behind Hikmet, so David is $10$ th.)

The runner in $8$ th place is Hikmet.

The answer is $\boxed{\textbf{(B)}\ \text{Hikmet}}$ .

Why this works

Clues like "3 places behind" link runners in a chain, but a chain needs one fixed point before any place is known. Find the clue that gives an exact place, then turn each "ahead by $k$ " into subtracting $k$ and each "behind by $k$ " into adding $k$ . The six unnamed runners just fill the empty places and can be ignored.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Only one clue gives an absolute position, so start there and follow the chain of relative clues. "Ahead" means a smaller place number, "behind" a larger one.

Marta is $6$ th.
Marta is $1$ behind Jack, so Jack is $5$ th.
Jack is $2$ behind Todd, so Todd is $3$ rd.
Todd is $1$ behind Rand, so Rand is $2$ nd.
Rand is $6$ ahead of Hikmet, so Hikmet is $2 + 6 = 8$ th.
David is $2$ behind Hikmet, so David is $10$ th.

The $8$ th-place finisher is Hikmet: $\boxed{\textbf{(B)}\ \text{Hikmet}}$ .

Review in the AoPS books

Intermediate Counting & Probability, Chapter 2: Sets and Logic (§2.4)